Suppose that we have two smooth functions from some real manifold $X$ into the complex plane $\mathbb C$:
$$f,g: X \to \mathbb C.$$
I'm vaguely wondering, if we treat $\mathbb C$ as a plane equipped with a flat metric, if there's a relationship between (complex) multiplication and the Frechet derivative map, i.e. a kind of "product rule" which relates $h(x):= f(x)g(x)$ to $f$ and $g$.
My super naive thinking would be that something like $(\ast)$ holds:
$$\phantom{(\ast)} \qquad D_x h = g(x)D_x f + f(x) D_x g \qquad (\ast)$$
(where $D_x f : T_x X \to T_{f(x)} \mathbb C$ denotes the Frechet derivative of $f$, for example),
but this doesn't even make sense, as $D_x f$, $D_x h$ and $D_x g$ can all land in different tangent spaces. This leads to my question.
Question: Is there any meaningful way to relate the derivative map of $h$ in terms of $f$ and $g$, given that there one can identify tangent spaces of $\mathbb C$ using isometries?
Failing that, is it possible to at least estimate the operator norm of $D_x h$ from that of $D_x f$ and $D_x g$?
Thanks for reading!
Since $\mathbb{C}$ is a vector space, its tangent spaces can be canonically identified with itself. Hence, for each $x\in X$, we may view $D_xf$ as a map $$D_xf:T_xX\to\mathbb{C}.$$ Then, $$D_x(fg)=f(x)D_xg+g(x)D_xf.$$ To prove this formula, note that $fg=m\circ(f\times g)$ where $m:\mathbb{C}\times\mathbb{C}\to\mathbb{C}$ is multiplication and $f\times g:X\to\mathbb{C}\times\mathbb{C}$ is the map $(f\times g)(x)=(f(x),g(x))$. Then, $D_x(f\times g)=D_xf\times D_xg:T_xX\to\mathbb{C}\times\mathbb{C}$. Moreover, after identifying $T_z\mathbb{C}$ with $\mathbb{C}$ for $z\in\mathbb{C}$ we have that for all $(z_1,z_2)\in\mathbb{C}\times\mathbb{C}$, $$D_{(z_1,z_2)}m:\mathbb{C}\times\mathbb{C}\to\mathbb{C},\quad (a,b)\mapsto z_1b+z_2a$$ since $$(D_{(z_1,z_2)}m)(a,b)=\left.\frac{d}{dt}\right|_{t=0}(z_1+at)(z_2+bt)=z_1b+z_2a.$$ Combining this with the chain rule, we get $$D_x(fg)=D_x(m\circ(f\times g))=D_{(f(x),g(x))}m\circ(D_xf\times D_xg)=f(x)D_xg+g(x)D_xf.$$